/* -*-  Mode:C; c-basic-offset:4; tab-width:4 -*-
 ****************************************************************************
 * (C) 2003 - Rolf Neugebauer - Intel Research Cambridge
 ****************************************************************************
 *
 *        File: math.c
 *      Author: Rolf Neugebauer (neugebar@dcs.gla.ac.uk)
 *     Changes:
 *
 *        Date: Aug 2003
 *
 * Environment: Xen Minimal OS
 * Description:  Library functions for 64bit arith and other
 *               from freebsd, files in sys/libkern/ (qdivrem.c, etc)
 *
 ****************************************************************************
 * $Id: math.c,v 1.1 2012/10/10 20:09:31 william Exp $
 ****************************************************************************
 *-
 * Copyright (c) 1992, 1993
 *      The Regents of the University of California.  All rights reserved.
 *
 * This software was developed by the Computer Systems Engineering group
 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
 * contributed to Berkeley.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions
 * are met:
 * 1. Redistributions of source code must retain the above copyright
 *    notice, this list of conditions and the following disclaimer.
 * 2. Redistributions in binary form must reproduce the above copyright
 *    notice, this list of conditions and the following disclaimer in the
 *    documentation and/or other materials provided with the distribution.
 * 3. All advertising materials mentioning features or use of this software
 *    must display the following acknowledgement:
 *      This product includes software developed by the University of
 *      California, Berkeley and its contributors.
 * 4. Neither the name of the University nor the names of its contributors
 *    may be used to endorse or promote products derived from this software
 *    without specific prior written permission.
 *
 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
 * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
 * SUCH DAMAGE.
 *
 * $FreeBSD: src/sys/libkern/divdi3.c,v 1.6 1999/08/28 00:46:31 peter Exp $
*/

#include "types.h"
#include "kernel.h"

/*
 * Depending on the desired operation, we view a `long long' (aka quad_t) in
 * one or more of the following formats.
 */
union uu {
        int64_t            q;              /* as a (signed) quad */
        int64_t            uq;             /* as an unsigned quad */
        long               sl[2];          /* as two signed longs */
        unsigned long      ul[2];          /* as two unsigned longs */
};
/* XXX RN: Yuck hardcoded endianess :) */
#define _QUAD_HIGHWORD 1
#define _QUAD_LOWWORD 0
/*
 * Define high and low longwords.
 */
#define H               _QUAD_HIGHWORD
#define L               _QUAD_LOWWORD

/*
 * Total number of bits in a quad_t and in the pieces that make it up.
 * These are used for shifting, and also below for halfword extraction
 * and assembly.
 */
#define CHAR_BIT        8               /* number of bits in a char */
#define QUAD_BITS       (sizeof(int64_t) * CHAR_BIT)
#define LONG_BITS       (sizeof(long) * CHAR_BIT)
#define HALF_BITS       (sizeof(long) * CHAR_BIT / 2)

/*
 * Extract high and low shortwords from longword, and move low shortword of
 * longword to upper half of long, i.e., produce the upper longword of
 * ((quad_t)(x) << (number_of_bits_in_long/2)).  (`x' must actually be u_long.)
 *
 * These are used in the multiply code, to split a longword into upper
 * and lower halves, and to reassemble a product as a quad_t, shifted left
 * (sizeof(long)*CHAR_BIT/2).
 */
#define HHALF(x)        ((x) >> HALF_BITS)
#define LHALF(x)        ((x) & ((1UL << HALF_BITS) - 1))
#define LHUP(x)         ((x) << HALF_BITS)

/*
 * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
 * section 4.3.1, pp. 257--259.
 */
#define B       (1UL << HALF_BITS)      /* digit base */

/* Combine two `digits' to make a single two-digit number. */
#define COMBINE(a, b) (((unsigned long)(a) << HALF_BITS) | (b))

/* select a type for digits in base B: use unsigned short if they fit */
/* #if ULONG_MAX == 0xffffffff && USHORT_MAX >= 0xffff
typedef unsigned short digit;
#else */
typedef unsigned long digit;
/* #endif */


/*
 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
 * `fall out' the left (there never will be any such anyway).
 * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
 */
static void
shl(register digit *p, register int len, register int sh)
{
        register int i;

        for (i = 0; i < len; i++)
                p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
        p[i] = LHALF(p[i] << sh);
}

/*
 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
 *
 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
 * fit within u_long.  As a consequence, the maximum length dividend and
 * divisor are 4 `digits' in this base (they are shorter if they have
 * leading zeros).
 */
uint64_t
__qdivrem(uint64_t uq, uint64_t vq, uint64_t *arq)
{
        union uu tmp;
        digit *u, *v, *q;
        register digit v1, v2;
        unsigned long qhat, rhat, t;
        int m, n, d, j, i;
        digit uspace[5], vspace[5], qspace[5];

        /*
         * Take care of special cases: divide by zero, and u < v.
         */
        if (vq == 0) {
                /* divide by zero. */
                static volatile const unsigned int zero = 0;

                tmp.ul[H] = tmp.ul[L] = 1 / zero;
                if (arq)
                        *arq = uq;
                return tmp.q;
        }
        if (uq < vq) {
                if (arq)
                        *arq = uq;
                return 0;
        }
        u = &uspace[0];
        v = &vspace[0];
        q = &qspace[0];

        /*
         * Break dividend and divisor into digits in base B, then
         * count leading zeros to determine m and n.  When done, we
         * will have:
         *      u = (u[1]u[2]...u[m+n]) sub B
         *      v = (v[1]v[2]...v[n]) sub B
         *      v[1] != 0
         *      1 < n <= 4 (if n = 1, we use a different division algorithm)
         *      m >= 0 (otherwise u < v, which we already checked)
         *      m + n = 4
         * and thus
         *      m = 4 - n <= 2
         */
        tmp.uq = uq;
        u[0] = 0;
        u[1] = HHALF(tmp.ul[H]);
        u[2] = LHALF(tmp.ul[H]);
        u[3] = HHALF(tmp.ul[L]);
        u[4] = LHALF(tmp.ul[L]);
        tmp.uq = vq;
        v[1] = HHALF(tmp.ul[H]);
        v[2] = LHALF(tmp.ul[H]);
        v[3] = HHALF(tmp.ul[L]);
        v[4] = LHALF(tmp.ul[L]);
        for (n = 4; v[1] == 0; v++) {
                if (--n == 1) {
                        unsigned long rbj;      /* r*B+u[j] (not root boy jim) */
                        digit q1, q2, q3, q4;

                        /*
                         * Change of plan, per exercise 16.
                         *      r = 0;
                         *      for j = 1..4:
                         *              q[j] = floor((r*B + u[j]) / v),
                         *              r = (r*B + u[j]) % v;
                         * We unroll this completely here.
                         */
                        t = v[2];       /* nonzero, by definition */
                        q1 = u[1] / t;
                        rbj = COMBINE(u[1] % t, u[2]);
                        q2 = rbj / t;
                        rbj = COMBINE(rbj % t, u[3]);
                        q3 = rbj / t;
                        rbj = COMBINE(rbj % t, u[4]);
                        q4 = rbj / t;
                        if (arq)
                                *arq = rbj % t;
                        tmp.ul[H] = COMBINE(q1, q2);
                        tmp.ul[L] = COMBINE(q3, q4);
                        return tmp.q;
                }
        }

        /*
         * By adjusting q once we determine m, we can guarantee that
         * there is a complete four-digit quotient at &qspace[1] when
         * we finally stop.
         */
        for (m = 4 - n; u[1] == 0; u++)
                m--;
        for (i = 4 - m; --i >= 0;)
                q[i] = 0;
        q += 4 - m;

        /*
         * Here we run Program D, translated from MIX to C and acquiring
         * a few minor changes.
         *
         * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
         */
        d = 0;
        for (t = v[1]; t < B / 2; t <<= 1)
                d++;
        if (d > 0) {
                shl(&u[0], m + n, d);           /* u <<= d */
                shl(&v[1], n - 1, d);           /* v <<= d */
        }
        /*
         * D2: j = 0.
         */
        j = 0;
        v1 = v[1];      /* for D3 -- note that v[1..n] are constant */
        v2 = v[2];      /* for D3 */
        do {
                register digit uj0, uj1, uj2;

                /*
                 * D3: Calculate qhat (\^q, in TeX notation).
                 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
                 * let rhat = (u[j]*B + u[j+1]) mod v[1].
                 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
                 * decrement qhat and increase rhat correspondingly.
                 * Note that if rhat >= B, v[2]*qhat < rhat*B.
                 */
                uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
                uj1 = u[j + 1]; /* for D3 only */
                uj2 = u[j + 2]; /* for D3 only */
                if (uj0 == v1) {
                        qhat = B;
                        rhat = uj1;
                        goto qhat_too_big;
                } else {
                        unsigned long nn = COMBINE(uj0, uj1);
                        qhat = nn / v1;
                        rhat = nn % v1;
                }
                while (v2 * qhat > COMBINE(rhat, uj2)) {
qhat_too_big:
                        qhat--;
                        if ((rhat += v1) >= B)
                                break;
                }
                /*
                 * D4: Multiply and subtract.
                 * The variable `t' holds any borrows across the loop.
                 * We split this up so that we do not require v[0] = 0,
                 * and to eliminate a final special case.
                 */
                for (t = 0, i = n; i > 0; i--) {
                        t = u[i + j] - v[i] * qhat - t;
                        u[i + j] = LHALF(t);
                        t = (B - HHALF(t)) & (B - 1);
                }
                t = u[j] - t;
                u[j] = LHALF(t);
                /*
                 * D5: test remainder.
                 * There is a borrow if and only if HHALF(t) is nonzero;
                 * in that (rare) case, qhat was too large (by exactly 1).
                 * Fix it by adding v[1..n] to u[j..j+n].
                 */
                if (HHALF(t)) {
                        qhat--;
                        for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
                                t += u[i + j] + v[i];
                                u[i + j] = LHALF(t);
                                t = HHALF(t);
                        }
                        u[j] = LHALF(u[j] + t);
                }
                q[j] = qhat;
        } while (++j <= m);             /* D7: loop on j. */

        /*
         * If caller wants the remainder, we have to calculate it as
         * u[m..m+n] >> d (this is at most n digits and thus fits in
         * u[m+1..m+n], but we may need more source digits).
         */
        if (arq) {
                if (d) {
                        for (i = m + n; i > m; --i)
                                u[i] = (u[i] >> d) |
                                       LHALF(u[i - 1] << (HALF_BITS - d));
                        u[i] = 0;
                }
                tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
                tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
                *arq = tmp.q;
        }

        tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
        tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
        return tmp.q;
}


/*
 * Divide two signed quads.
 * ??? if -1/2 should produce -1 on this machine, this code is wrong
 */
int64_t __divdi3(int64_t a, int64_t b)
{
        uint64_t ua, ub, uq;
        int neg;

        if (a < 0)
                ua = -(uint64_t)a, neg = 1;
        else
                ua = a, neg = 0;
        if (b < 0)
                ub = -(uint64_t)b, neg ^= 1;
        else
                ub = b;
        uq = __qdivrem(ua, ub, (uint64_t *)0);
        return (neg ? -uq : uq);
}

/*
 * Divide two unsigned quads.
 */
uint64_t
__udivdi3(uint64_t a, uint64_t b)
{
        return __qdivrem(a, b, (uint64_t *)0);
}


/*
 * Return remainder after dividing two unsigned quads.
 */
uint64_t
__umoddi3(uint64_t a, uint64_t b)
{
        uint64_t r;

        (void)__qdivrem(a, b, &r);
        return r;
}

/*
 * Return ceil(log_2(x))
 * We shift our input right until we get zero. The number of times we had to
 * shift before getting zero gives us the ceiling of log2(x), except for powers
 * of 2, in which case it gives us 1 + log2(x). Thus, we check whether it's a
 * power of two, and special case that.
 * author: dap
 */
int log2(int x)
{
        int current = x;
        /* y keeps track of 2^(result) to see if our input was a power of 2 */
        int y = 1;
        int result = 0;
        while (current) {
                current >>= 1;
                ++result;
                y <<= 1;
        }
        y >>= 1;
        if (y == x)
                return result - 1;

        return result;
}
